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Mathematics 12 Online
OpenStudy (anonymous):

find all the zeroes in the equation. -3x^4+27x^2+1200=0

OpenStudy (denisaboichuk):

10

OpenStudy (witty_roger):

very easy...substitute x^2 as some variable... say p...den solve the quadratic equation in p....-3p^2+27p+1200=0

OpenStudy (witty_roger):

so 3p^2-27p-1200=0.....divide throughout by 3 p^2-9p-400=0............solving my factorization p^2-25p+16p-400=0 p(p-25)+16(p-25)=0 so (p-25)(p+16)=0 this means Either p-25=0 OR p+16=0 which means either p=25 OR p=-16 but p=x^2 so x=+5 or -5 OR x=+4i or -4i

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